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Since the block are on a frictionless surface, they don't interact with the Earth, so it doesn't make much sense to talk about the "block-earth" system. Within the context of this block-block system, the mechanical energy is always conserved. In terms of equations, at the point of collision we have $m_{1}v_{10}=m_{1}v_{1}+m_{2}v_{2}$ ...

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First of all, I agree with the result you obtained using conversation of energy. Your logic with the Work-Energy Theorem does not work because the total work done on block A is not $-1/2kx_0^2$ (I am denoting the max compression by $x_0$). Suppose that block A first comes into contact with the spring at $s=0$ and suppose that its velocity reaches $5.5\, ... 0 Have a look at the two animated gifs at the top of the Brownian Motion page on Wikipedia. Essentially your 2mm sphere will follow a random path due to the random collisions from the air molecules that given enough time will travel to most places in the container (assuming a starting velocity of 0). If you give an initial velocity then its probably ... 0 You always have to satisfy the momentum equations, which is only the linear momentum equation for this one dimensional case: $$m_1v(t_1)-m_1v(t_0) = \int_{t_0}^{t_1} F dt$$ Assume the collision is completely elastic and all is conservative, so no plastic deformation, drag or any kind of damping. Then the only force which acts is the gravity: $$F = ... 0 Neutron stars are a special case. When the neutron stars collide a black hole is formed within milliseconds. A gamma ray burst of less than 2 seconds is expected to be observable from a great distant, as well as gravitational waves. About 1% of a solar mass is expected to be ejected and include heavy elements. On June 3rd 2013 the Swift gamma ray ... 2 The only thing that the device "knows" when it is hit, is the force with which it gets hit, and the duration of that hit. Transfer of momentum m\Delta v = F\Delta t. So what matters is the momentum of the hammer's head - or more specifically, the momentum that you are able to transfer. Ultimately it comes down to giving the most momentum to the head of the ... 1 Let S and S' be the two inertial frame and S' moving with a constant velocity v w.r.t. S frame. Now a force F acting on the particle at point A and displace it to the point B. If the position x-coordinates of A point and B point in S' frame are (x_1',x_2') and in S frame are (x_1,x_2) then at any time t, x_1=x_1'+vt and ... 1 It came from the wall. If a ball hits a wall at constant velocity, then it's not going to "slow down and eventually come to rest". It's going to bounce back, and in fact, if the collision is elastic, it won't even necessarily slow down. The force on the ball from the wall serves to change the direction of the velocity (and possibly decrease its magnitude ... 1 Suppose total kinetic energy were conserved (i.e. if you can ignore potential energy changes), then because it's a nonlinear fuunction of the velocities, the sum of the kinetic energies of the particles that make up a composite body does not equal the kinetic energy of the center the center of mass of the object. In practice, what this means is that you ... 4 I'm 27 and since I was about 15 I had the same doubt you do. Only a couple of years ago I realized why momentum is always conserved in a collision, whereas the same is not enforced for energy. (They must have told me this at some point -- or points --, but I guess sometimes I just don't pay much attention) First of all, let's make it empirically clear that ... 5 Possibly your confusion arises from not considering that KE is a scalar, whilst momentum is a vector. Yes, there is of course a connection between KE and momentum: K = p^2/2m (for non-relativistic bodies). Thus, for two equal-mass particles heading directly towards each other at equal speeds v, their velocities are \pm {\bf v} and their momenta \pm ... 4 Is there no relation between the momentum and kinetic energy? Are they not linked with each other? No, they're not. Conservation of momentum only says that$$m_1 v_{1_-} + m_2 v_{2_-} = m_1 v_{1_+} + m_2 v_{2_+}$$I used subscripts 1 and 2 denote the particles involved in the collision and the subscripts - and + denote the pre- and post-collision ... 0 Individual momenta are not conserved, only the total momentum is. It is not entirely determined with the kinetic energy, there are some additional "degrees of freedom" or "dimensions", loosely speaking. The total energy is conserved too, you just have to consider a potential energy of interaction during (even elastic) collision. Heat released in a body is ... 1 Consider the relative velocity of one body with respect to another. That's all. So, relative velocity of approach would mean the velocity of body 1 coming towards body 2, signified by v_1-v_2 And the relative velocity of separation would mean the velocity of body 1 going away from body 2, signified by u_1-u_2 as seen by body 2, in both the cases Yes, ... 1 The kinetic energy of the center of masses of the two colliding bodies may increase if their internal structure changes, i.e. if at least one of the bodies were in an excited state, and the conservation laws of energy and momentum allow the exceeding energy to transform into kinetic energy. 1 As some other materials in addition of those definitely helpful references proposed by Michiel, I found the following very useful resources: a full review of many concepts in droplet dynamics has been presented in this book: Ashgriz, N. Handbook of Atomization and Sprays. Vol. 11. Springer New York, 2010. for droplet collision: Rein, Martin. ... 0 It depends. If a projectile is scattered elastically from a target, it does not change the target state. If the state changes, the projectile may loose or gain energy. For example, if you scatter from an atom in the ground state, only excitations of atom are possible; they need energy for that. But if you scatter from an excites atomic state, the final atom ... -3 No collision is possible in between two electrons because the electron carry negative charge and push each other away with some assumption 1 Books For fundamentals I prefer books over papers, because they are typically more thorough and a little bit more 'slow' in the introduction of concepts. There are many books that will cover, some of, the topics that you mention. I will mention below the 3 books that where most useful to me in the past. 1) An excellent resource for a theoretical foundation ... 2 You're right, it is fairly obvious. Conservation of linear momentum applies at all times! If an object is irradiated by alpha particles, each with mass \sim 4m_u and travelling with velocity v, and if it absorbs N of these particles in a time t. Then the force exerted on the object is the rate of change of momentum.$$ F = \frac{4Nm_u v}{t}$\$ Of ...

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In one way, we can assume that particles repulsing when they "approach close by" is similar to particles with larger sizes colliding. It means that particles cannot approach arbitrarily close. This may not be exactly what you are distinguishing with "physical collision", but it has consequences for ideal gases. The greater the distance of this ...

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An ideal gas should consist of pointlike particles that are non-interacting, except if they collide, in which case they should do so elastically, without losing kinetic energy. I do not think there is any distinction here between a collision and a repulsive force. Any short-range repulsive force between particles (short-range compared with the average ...

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Consider a plasma that just's been formed and then left alone. Being far from equilibrium, the plasma will evolve towards an equilibrium state. At this stage, it's not very useful to characterize the plasma with a temperature because the velocity distribution would bear little resemblance to a Boltzmann distribution, or really any kind of distribution ...

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There are lots of different types of plasmas. In a thermal plasma the electrons and ions will have the same temperature. In a non-thermal plasma the discharge is driven by some external power supply e.g. capacitatively coupled RF, inductively coupled, pulsed DC E field etc. In a non-thermal plasma the electrons generally have a higher temperature than ...

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If the string is massless and taut, then the wave velocity is infinite - that is, a component of the force at one mass will immediately be felt at the other mass. But to answer your first question, not all the force of the impact will be transmitted along the string, as made clear by this diagram: As for your second question: momentum for the system is ...

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